If C(16,r) = C(16,r+2), then find r. Explain how you know.











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Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?










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    up vote
    2
    down vote

    favorite












    Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?










    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?










      share|cite|improve this question













      Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?







      combinations






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      asked 2 hours ago









      Gold Pony Boy

      152




      152






















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          Note that $binom{n}{k} = binom{n}{n-k}.$



          What equation does this give for $r$ in your scenario? What is the solution to that equation?



          Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.






          share|cite|improve this answer























          • Your equation does not imply $$binom na=binom nbimplies a=n-b$$
            – Kemono Chen
            1 hour ago












          • @KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
            – Display name
            1 hour ago












          • Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
            – Kemono Chen
            1 hour ago


















          up vote
          3
          down vote













          Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?






          share|cite|improve this answer





















          • Okay, I see the connection now, so r must be 7...?
            – Gold Pony Boy
            1 hour ago










          • @GoldPonyBoy Yes.
            – Carl Schildkraut
            34 mins ago


















          up vote
          0
          down vote













          $C_r^{16}=C_{r+2}^{16}$
          Then using the definition of combination you'll get



          $$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
          $$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
          $$implies (16-r)(15-r)=(r+2)(r+1)$$
          $$implies 240-31r+r^2=r^2+3r+2$$
          $$implies 34r=238$$
          $$implies r=7$$






          share|cite|improve this answer























          • Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
            – Gold Pony Boy
            1 hour ago






          • 2




            @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
            – mathematics2x2life
            1 hour ago










          • @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
            – Gold Pony Boy
            1 hour ago












          • @mathematics2x2life Actually wait,where did the r! from the left side go after the first step?
            – Gold Pony Boy
            1 hour ago






          • 1




            @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
            – Fareed AF
            1 hour ago











          Your Answer





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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          4
          down vote



          accepted










          Note that $binom{n}{k} = binom{n}{n-k}.$



          What equation does this give for $r$ in your scenario? What is the solution to that equation?



          Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.






          share|cite|improve this answer























          • Your equation does not imply $$binom na=binom nbimplies a=n-b$$
            – Kemono Chen
            1 hour ago












          • @KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
            – Display name
            1 hour ago












          • Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
            – Kemono Chen
            1 hour ago















          up vote
          4
          down vote



          accepted










          Note that $binom{n}{k} = binom{n}{n-k}.$



          What equation does this give for $r$ in your scenario? What is the solution to that equation?



          Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.






          share|cite|improve this answer























          • Your equation does not imply $$binom na=binom nbimplies a=n-b$$
            – Kemono Chen
            1 hour ago












          • @KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
            – Display name
            1 hour ago












          • Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
            – Kemono Chen
            1 hour ago













          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          Note that $binom{n}{k} = binom{n}{n-k}.$



          What equation does this give for $r$ in your scenario? What is the solution to that equation?



          Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.






          share|cite|improve this answer














          Note that $binom{n}{k} = binom{n}{n-k}.$



          What equation does this give for $r$ in your scenario? What is the solution to that equation?



          Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 5 mins ago

























          answered 2 hours ago









          Display name

          789313




          789313












          • Your equation does not imply $$binom na=binom nbimplies a=n-b$$
            – Kemono Chen
            1 hour ago












          • @KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
            – Display name
            1 hour ago












          • Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
            – Kemono Chen
            1 hour ago


















          • Your equation does not imply $$binom na=binom nbimplies a=n-b$$
            – Kemono Chen
            1 hour ago












          • @KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
            – Display name
            1 hour ago












          • Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
            – Kemono Chen
            1 hour ago
















          Your equation does not imply $$binom na=binom nbimplies a=n-b$$
          – Kemono Chen
          1 hour ago






          Your equation does not imply $$binom na=binom nbimplies a=n-b$$
          – Kemono Chen
          1 hour ago














          @KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
          – Display name
          1 hour ago






          @KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
          – Display name
          1 hour ago














          Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
          – Kemono Chen
          1 hour ago




          Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
          – Kemono Chen
          1 hour ago










          up vote
          3
          down vote













          Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?






          share|cite|improve this answer





















          • Okay, I see the connection now, so r must be 7...?
            – Gold Pony Boy
            1 hour ago










          • @GoldPonyBoy Yes.
            – Carl Schildkraut
            34 mins ago















          up vote
          3
          down vote













          Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?






          share|cite|improve this answer





















          • Okay, I see the connection now, so r must be 7...?
            – Gold Pony Boy
            1 hour ago










          • @GoldPonyBoy Yes.
            – Carl Schildkraut
            34 mins ago













          up vote
          3
          down vote










          up vote
          3
          down vote









          Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?






          share|cite|improve this answer












          Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Carl Schildkraut

          10.6k11438




          10.6k11438












          • Okay, I see the connection now, so r must be 7...?
            – Gold Pony Boy
            1 hour ago










          • @GoldPonyBoy Yes.
            – Carl Schildkraut
            34 mins ago


















          • Okay, I see the connection now, so r must be 7...?
            – Gold Pony Boy
            1 hour ago










          • @GoldPonyBoy Yes.
            – Carl Schildkraut
            34 mins ago
















          Okay, I see the connection now, so r must be 7...?
          – Gold Pony Boy
          1 hour ago




          Okay, I see the connection now, so r must be 7...?
          – Gold Pony Boy
          1 hour ago












          @GoldPonyBoy Yes.
          – Carl Schildkraut
          34 mins ago




          @GoldPonyBoy Yes.
          – Carl Schildkraut
          34 mins ago










          up vote
          0
          down vote













          $C_r^{16}=C_{r+2}^{16}$
          Then using the definition of combination you'll get



          $$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
          $$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
          $$implies (16-r)(15-r)=(r+2)(r+1)$$
          $$implies 240-31r+r^2=r^2+3r+2$$
          $$implies 34r=238$$
          $$implies r=7$$






          share|cite|improve this answer























          • Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
            – Gold Pony Boy
            1 hour ago






          • 2




            @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
            – mathematics2x2life
            1 hour ago










          • @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
            – Gold Pony Boy
            1 hour ago












          • @mathematics2x2life Actually wait,where did the r! from the left side go after the first step?
            – Gold Pony Boy
            1 hour ago






          • 1




            @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
            – Fareed AF
            1 hour ago















          up vote
          0
          down vote













          $C_r^{16}=C_{r+2}^{16}$
          Then using the definition of combination you'll get



          $$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
          $$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
          $$implies (16-r)(15-r)=(r+2)(r+1)$$
          $$implies 240-31r+r^2=r^2+3r+2$$
          $$implies 34r=238$$
          $$implies r=7$$






          share|cite|improve this answer























          • Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
            – Gold Pony Boy
            1 hour ago






          • 2




            @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
            – mathematics2x2life
            1 hour ago










          • @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
            – Gold Pony Boy
            1 hour ago












          • @mathematics2x2life Actually wait,where did the r! from the left side go after the first step?
            – Gold Pony Boy
            1 hour ago






          • 1




            @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
            – Fareed AF
            1 hour ago













          up vote
          0
          down vote










          up vote
          0
          down vote









          $C_r^{16}=C_{r+2}^{16}$
          Then using the definition of combination you'll get



          $$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
          $$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
          $$implies (16-r)(15-r)=(r+2)(r+1)$$
          $$implies 240-31r+r^2=r^2+3r+2$$
          $$implies 34r=238$$
          $$implies r=7$$






          share|cite|improve this answer














          $C_r^{16}=C_{r+2}^{16}$
          Then using the definition of combination you'll get



          $$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
          $$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
          $$implies (16-r)(15-r)=(r+2)(r+1)$$
          $$implies 240-31r+r^2=r^2+3r+2$$
          $$implies 34r=238$$
          $$implies r=7$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          Fareed AF

          36711




          36711












          • Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
            – Gold Pony Boy
            1 hour ago






          • 2




            @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
            – mathematics2x2life
            1 hour ago










          • @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
            – Gold Pony Boy
            1 hour ago












          • @mathematics2x2life Actually wait,where did the r! from the left side go after the first step?
            – Gold Pony Boy
            1 hour ago






          • 1




            @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
            – Fareed AF
            1 hour ago


















          • Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
            – Gold Pony Boy
            1 hour ago






          • 2




            @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
            – mathematics2x2life
            1 hour ago










          • @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
            – Gold Pony Boy
            1 hour ago












          • @mathematics2x2life Actually wait,where did the r! from the left side go after the first step?
            – Gold Pony Boy
            1 hour ago






          • 1




            @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
            – Fareed AF
            1 hour ago
















          Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
          – Gold Pony Boy
          1 hour ago




          Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
          – Gold Pony Boy
          1 hour ago




          2




          2




          @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
          – mathematics2x2life
          1 hour ago




          @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
          – mathematics2x2life
          1 hour ago












          @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
          – Gold Pony Boy
          1 hour ago






          @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
          – Gold Pony Boy
          1 hour ago














          @mathematics2x2life Actually wait,where did the r! from the left side go after the first step?
          – Gold Pony Boy
          1 hour ago




          @mathematics2x2life Actually wait,where did the r! from the left side go after the first step?
          – Gold Pony Boy
          1 hour ago




          1




          1




          @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
          – Fareed AF
          1 hour ago




          @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
          – Fareed AF
          1 hour ago


















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